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I'm trying to factor the following polynomial by hand:

$-x^3 + 9x^2 - 24x + 20 = 0$

The simplest I could get is:

$-x^2(x-9) - 4(5x+5) = 0$

Any ideas on how I could go ahead and solve this by hand? This seems pretty tough.

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Did you try the rational zero theorem to see if they are any? Possbile candidates would be 1,2,4,5,10 and 20 ,with their opposites included. Give it a try. Once you have one, you can divide out its factor and you are left with a quadratic –  imranfat Apr 26 '13 at 19:35
    
Just by luck, I managed to guess right with $-x^3+5x^2+4x^2-24x+20=-x^2(x-5)+4(x-5)(x-1)$. Still, I think DonAntonio's approach is the way to go here. –  Mike Apr 26 '13 at 19:42
    
Here is a detailed solution. –  Mhenni Benghorbal Apr 26 '13 at 19:45

4 Answers 4

up vote 2 down vote accepted

You have not a polynomial, rather an equation, whose LHS is the cubic polynomial $$-x^{3}+9x^{2}-24x+20.\tag{0}$$

To get rid of the negative coefficient of $x^3$ multiply $(0)$ by $-1$ $$ \begin{equation*} -\left( -x^{3}+9x^{2}-24x+20\right) =x^{3}-9x^{2}+24x-20.\tag{1} \end{equation*} $$

By trial and error$^1$, inspection, or per DonAntonio's hint (Rational root test), we find that $x=5$ is a root: $5^{3}-9\left( 5\right) ^{2}+24\left( 5\right) -20=0$. As a consequence we can divide $(1)$ by $x-5$, using polynomial long division or Ruffini's rule to obtain a quadratic polynomial, since the remainder must be zero $$ \begin{equation*} \frac{x^{3}-9x^{2}+24x-20}{x-5}= x^{2}-4x+4 =\left( x-2\right)\tag{2} ^{2}. \end{equation*} $$

So $$ \begin{equation*} -x^{3}+9x^{2}-24x+20=-\left( x-5\right) \left( x-2\right) ^{2},\tag{3} \end{equation*} $$ whose roots are $x=5$ (single root) and $x=2$ (double root).

If you start with the root $x=2$, you get the same result:

$$\frac{x^{3}-9x^{2}+24x-20}{x-2}=x^{2}-7x+10=\left( x-2\right) \left( x-5\right).\tag{2'} $$

$^1$ Trial and error is a fundamental method of solving problems.

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Hint:

If $\,\frac{r}{s}\;,\;\;r,s\in\Bbb Z\,$ , is a rational root of the integer polynmomial $\,a_nx^n+\ldots+a_1x+a_0\;$ , then

$\,r\,\mid\,a_0\;,\;\;s\,\mid\,a_n\,$

Here, the divisors of $\,20\,$ are $\,\pm\{1,2,4,5,10,20\}\,$ ....

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1  
+1, concise and the most helpful hint. –  Américo Tavares Apr 26 '13 at 20:55

$x^3-9x^2+24x-20=0$

Using Cardano's method, put $x=t-\frac{-9}{3}=t+3$

$(t+3)^3-9(t+3)^2+24(t+3)-20=0$

$\implies t^3+27t+27-81-54t+24t+72-20=0$

$\implies t^3-3t-2=0$

Now we can follow standard Cardano method

or by observation , $t=-1\implies x=t+3=2$

Now, divide by $x-2$ to get a Quadratic

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Polynomial long division supports arbitrary denominators. Certain real polynomials are irreducible, but not of the form $x-c, c \in \mathbb{R}$. This is because the real numbers are not algebraically closed, which is why you need the complex numbers. However, we know all these irreducible polynomials are quadratics $ax^2+bx+c$, and studying the polynomial ring of the real numbers makes this pretty easy to prove. So if you tried dividing by one of those polynomials with those placeholder constants and setting your remainder to 0 you would find all the factors are all the combinations of a, b, and c that solve the equation.

Polynomial long division works like normal long division, where you take 121 and divide by first subtracting 10*10, then subtracting 2*10, and get a remainder of 1, except you use distributivity to say $x(x^2+2x) = x^3+2x^2$, so that you can line your first denominator term's highest term up with the highest-order term of the numerator. Then as you try and cancel out the result further, you would say $x(x^2+2x)+5(x^2+2x)=(x+5)(x^2+2x)$, so that you're saying $x^2+2x$ goes into the numerator $x+5$ times.

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